Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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40) please see attached
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**Chemical Equilibrium Constant Calculation**
Consider the equilibrium reaction:
\[ 2N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g) \]
In a particular experiment, the equilibrium concentration of N₂ is 0.048 M, of O₂ is 0.093 M, and of N₂O is 6.55 x 10⁻²¹ M. What is the value of the equilibrium constant \( K_c \)?
**Possible answers**:
a. \( 1.5 \times 10^{-18} \)
b. \( 2.0 \times 10^{-37} \)
c. \( 2.2 \times 10^{-36} \)
d. \( 3.1 \times 10^{-17} \)
e. \( 5.0 \times 10^{36} \)
---
Interpretation:
To determine the equilibrium constant \( K_c \), use the expression for the equilibrium constant \( K_c \) for the given balanced chemical equation:
\[ K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \]
From the given data:
\[ [N_2] = 0.048 \, M \]
\[ [O_2] = 0.093 \, M \]
\[ [N_2O] = 6.55 \times 10^{-21} \, M \]
Substitute these values into the expression for \( K_c \):
\[ K_c = \frac{(6.55 \times 10^{-21})^2}{(0.048)^2 \times (0.093)} \]
Calculate the numerator:
\[ (6.55 \times 10^{-21})^2 = 4.29 \times 10^{-41} \]
Calculate the denominator:
\[ (0.048)^2 \times 0.093 = 2.14 \times 10^{-4} \]
Finally, divide the numerator by the denominator to find \( K_c \):
\[ K_c = \frac{4.29 \times 10^{-41}}{2.14 \times 10^{-4}} = 2.0 \times 10^{-37} \]
Therefore, the correct answer is:
b. \( 2.0 \times 10^{-37}](https://content.bartleby.com/qna-images/question/48deb947-3d41-43de-ac43-bef03e01cc59/7c05f84d-01a3-40b9-8875-df4b65230bb9/h9pu4th.png)
Transcribed Image Text:---
**Chemical Equilibrium Constant Calculation**
Consider the equilibrium reaction:
\[ 2N_2(g) + O_2(g) \rightleftharpoons 2N_2O(g) \]
In a particular experiment, the equilibrium concentration of N₂ is 0.048 M, of O₂ is 0.093 M, and of N₂O is 6.55 x 10⁻²¹ M. What is the value of the equilibrium constant \( K_c \)?
**Possible answers**:
a. \( 1.5 \times 10^{-18} \)
b. \( 2.0 \times 10^{-37} \)
c. \( 2.2 \times 10^{-36} \)
d. \( 3.1 \times 10^{-17} \)
e. \( 5.0 \times 10^{36} \)
---
Interpretation:
To determine the equilibrium constant \( K_c \), use the expression for the equilibrium constant \( K_c \) for the given balanced chemical equation:
\[ K_c = \frac{[N_2O]^2}{[N_2]^2 [O_2]} \]
From the given data:
\[ [N_2] = 0.048 \, M \]
\[ [O_2] = 0.093 \, M \]
\[ [N_2O] = 6.55 \times 10^{-21} \, M \]
Substitute these values into the expression for \( K_c \):
\[ K_c = \frac{(6.55 \times 10^{-21})^2}{(0.048)^2 \times (0.093)} \]
Calculate the numerator:
\[ (6.55 \times 10^{-21})^2 = 4.29 \times 10^{-41} \]
Calculate the denominator:
\[ (0.048)^2 \times 0.093 = 2.14 \times 10^{-4} \]
Finally, divide the numerator by the denominator to find \( K_c \):
\[ K_c = \frac{4.29 \times 10^{-41}}{2.14 \times 10^{-4}} = 2.0 \times 10^{-37} \]
Therefore, the correct answer is:
b. \( 2.0 \times 10^{-37}
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